91影视

The sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$and the point $(伪,尾,纬)$

The goal is to determine the equation of the plane tangent to the sphere at a specific position. The equation of the plane tangent to the sphere is obtained by finding the normal of the surface of the sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$at the given point $(伪,尾,纬)$

The normal of the sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$at the point $(伪,尾,纬)$is:

$$\begin{gathered} {\left.鈭�f(x,y,z)\right|}_{(伪,尾,y)}=\left(2\right(x-a)\mathbf{i}+2(y-b)\mathbf{j}+2(y-c)\mathbf{k}{)}_{(a,尾,y)} \\ =(2(伪-a)\mathbf{i}+2(尾-b)\mathbf{j}+2(纬-c)\mathbf{k}) \\ =鉄�2(伪-a),2(尾-b),2(纬-c)鉄� \\ =鉄�(伪-a),(尾-b),(纬-c)鉄� \end{gathered}$$

The equation of the plane tangent to the sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$and the point $(伪,尾,纬)$is:

$(伪-a)(x-伪)+(尾-b)(y-尾)+(纬-c)(z-纬)=0$

The equation of the plane tangent to the sphere $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$and at the point $(伪,尾,纬)$is $(伪-a)(x-伪)+(尾-b)(y-尾)+(纬-c)(z-纬)=0$